Here's my go at Puzzle 16: Let's say we have two monotone functions \\(f : A\to B\\) and \\(g:B\to A\\) between preorders and we're wondering whether the following three conditions on \\(f\\) and \\(g\\) are equivalent:

$$\text{For all }a\text{ and }b,\ f(a)\leq b \iff a \leq g(b)$$

$$\text{For all }a,\ f(a)\text{ is the smallest $b$ with }a\leq g(b).$$

$$\text{For all }b,\ g(b)\text{ is the largest $a$ with }f(a)\leq b.$$

We'll show that the first and second are equivalent. First, note that since \\(g\\) is monotone, for any choice of \\(a\\) the set of all \\(b\\) such that \\(a\leq g(b)\\) is an upper set of \\(B\\). Therefore saying that \\(f(a)\\) is the smallest \\(b\\) with \\(a\leq g(b)\\) is saying that this upper set *is* the set of all elements of \\(B\\) at least as large as \\(f(a)\\). In other words, \\(a\leq g(b)\\) if and only if \\(b\\) is in this upper set if and only if \\(b \geq f(a)\\).

The equivalence between the first and third conditions is similar. I was surprised that you didn't need both the second and third to get something equivalent to the first! In fact, the second and third are already equivalent to each other.

$$\text{For all }a\text{ and }b,\ f(a)\leq b \iff a \leq g(b)$$

$$\text{For all }a,\ f(a)\text{ is the smallest $b$ with }a\leq g(b).$$

$$\text{For all }b,\ g(b)\text{ is the largest $a$ with }f(a)\leq b.$$

We'll show that the first and second are equivalent. First, note that since \\(g\\) is monotone, for any choice of \\(a\\) the set of all \\(b\\) such that \\(a\leq g(b)\\) is an upper set of \\(B\\). Therefore saying that \\(f(a)\\) is the smallest \\(b\\) with \\(a\leq g(b)\\) is saying that this upper set *is* the set of all elements of \\(B\\) at least as large as \\(f(a)\\). In other words, \\(a\leq g(b)\\) if and only if \\(b\\) is in this upper set if and only if \\(b \geq f(a)\\).

The equivalence between the first and third conditions is similar. I was surprised that you didn't need both the second and third to get something equivalent to the first! In fact, the second and third are already equivalent to each other.